The stress-energy tensor of a perfect fluid is given by
$$T^{\mu\nu}=\left(\rho+pc^{-2}\right)u^\mu u^\nu+pg^{\mu\nu}$$
The divergence of the stress-energy tensor is zero: $\nabla_\mu T^{\mu\nu}=0$. Hence
$$\nabla_\mu\left(\rho+pc^{-2}\right)u^\mu u^\nu+\nabla_\mu pg^{\mu\nu}=0$$
Expanding the first term, and using the product rule on the second term, yields
$$\nabla_\mu\rho u^\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$
Using the product rule once again on the first term yields
$$\left(\nabla_\mu\rho u^\mu\right) u^\nu+\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$
By the continuity equation, $\nabla_\mu\rho u^\mu=0$. Hence
$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$
The divergence of the metric tensor is zero: $\nabla_\mu g^{\mu\nu}=0$. Hence
$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}=0$$
Finally, using tensor contraction on the last term yields
$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\nabla^{\nu}p=0$$
We now turn to the Cauchy momentum equation in the Euler equations:
$$0=\rho\left(\frac{\partial}{\partial t}+\vec{u}\cdot\vec{\nabla}\right)\vec{u}+\vec{\nabla}p= \rho\left(c\nabla_0+\vec{u}\cdot\vec{\nabla}\right)\vec{u}+\vec{\nabla}p$$
Using the non-relativistic approximation $\gamma\approx1$ we obtain:
$$0\approx \rho\left(\gamma c\nabla_0+\gamma\vec{u}\cdot\vec{\nabla}\right)\gamma\vec{u}+\vec{\nabla}p= \rho u^{\mu}\nabla_{\mu}u^i+\nabla^ip$$
Compare this with the result obtained from the stress-energy tensor:
$$0=\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\nabla^{\nu}p$$
Why is there an extra term ($\nabla_\mu pc^{-2}u^\mu u^\nu$)? Does it vanish in the non-relativistic limit, simply because of the $c^{-2}$ factor?